Limit, Continuity and Differentiability — JEE Main Maths PYQ MCQs with Solutions

Solution: LHS $=\lim_{x\to1}(x^3+x^2+x+1)=4$. RHS: $\dfrac{x^3-k^3}{x^2-k^2}=\dfrac{x^2+xk+k^2}{x+k}\to\dfrac{3k^2}{2k}=\dfrac{3k}{2}$ as $x\to k$. Setting $\dfrac{3k}{2}=4$ gives $k=\dfrac{8}{3}$.

Solution: Using $\cot\theta\sim\dfrac{1}{\theta}$ and $\sin x\sim x$ for small $x$: numerator $\sim x\cdot\dfrac{1}{4x}=\dfrac14$; denominator $\sim x^2\cdot\dfrac{1}{4x^2}=\dfrac14$. Ratio $\to 1$.

Solution: $1-\cos2x=2\sin^2x\approx2x^2$, $(3+\cos x)\to4$, $\tan4x\approx4x$. So the expression $\approx\dfrac{2x^2\cdot4}{x\cdot4x}=2$.

Solution: $\sqrt{\tfrac12(1-\cos^2x)}=\sqrt{\tfrac12\sin^2x}=\dfrac{|\sin x|}{\sqrt2}$. As $x\to0^+$, the expression $\to\dfrac{1}{\sqrt2}$; as $x\to0^-$, it $\to-\dfrac{1}{\sqrt2}$. The two-sided limit does not exist, so the answer is None of these.

Solution: The sum is $\dfrac{1+2+\cdots+n}{1+n^2}=\dfrac{n(n+1)/2}{1+n^2}$. Dividing numerator and denominator by $n^2$ and letting $n\to\infty$ gives $\dfrac12$.

Solution: This limit is simply $G'(1)$. Since $G'(x)=\dfrac{-x}{\sqrt{25-x^2}}$, we get $G'(1)=\dfrac{-1}{\sqrt{24}}$.

Solution: $\left(\dfrac{x-3}{x+2}\right)^x=\left(1-\dfrac{5}{x+2}\right)^x\to e^{-5}$.

Solution: For $x$ close to $0$ (but $x \neq 0$), $0 < \tan^2 x < 1$, so $[\tan^2 x] = 0$, and $f(0) = [\tan^2 0] = 0$. Thus $\lim_{x\to 0} f(x) = 0 = f(0)$, so $f$ is continuous at $x = 0$.

Solution: Using $\lim_{t\to 0} \dfrac{\log(1+t)}{t} = 1$: $\lim_{x\to 0} \dfrac{\log(1+ax)}{x} = a$ and $\lim_{x\to 0} \dfrac{-\log(1-bx)}{x} = b$. So the limit of $f(x)$ as $x\to 0$ is $a+b$, which must be assigned as $f(0)$.

Solution: $|x|$ itself has a corner at $x=0$. In addition, $\big||x|-1\big|$ has corners wherever its inner expression $|x|-1$ vanishes, i.e. at $x=\pm1$. So $f$ is not differentiable at $x=-1,0,1$.

Solution: The absolute value function is continuous everywhere, so $g=|f|$ is a composition of continuous functions whenever $f$ is continuous, hence continuous. Ontoness and one-oneness need not transfer (e.g. $f(x)=x$ is onto and one-one, but $|x|$ is neither), and differentiability of $f$ does not guarantee differentiability of $|f|$ at points where $f$ vanishes.

Solution: Since $g$ is the inverse of $f$, $g'(x)=\dfrac{1}{f'(g(x))}=1+\{g(x)\}^5$.

Solution: $y=\sec(\tan^{-1}x)=\sqrt{1+x^2}$, so $\dfrac{dy}{dx}=\dfrac{x}{\sqrt{1+x^2}}$. At $x=1$: $\dfrac{1}{\sqrt2}$.

Solution: $\cot^3x-\tan x=\dfrac{\cos^4x-\sin^4x}{\sin^3x\cos x}=\dfrac{\cos2x}{\sin^3x\cos x}$. Also $\cos\left(x+\frac{\pi}{4}\right)=\dfrac{\cos x-\sin x}{\sqrt2}$ and $\cos2x=(\cos x-\sin x)(\cos x+\sin x)$. Cancelling $(\cos x-\sin x)$ gives $\dfrac{\sqrt2(\cos x+\sin x)}{\sin^3x\cos x}$, which at $x=\pi/4$ (where $\sin x=\cos x=1/\sqrt2$) equals $\dfrac{\sqrt2\cdot\sqrt2}{1/4}=8$.

Solution: Let $t=y^4\to0^+$. $\sqrt{1+t}\approx1+t/2$, so $1+\sqrt{1+t}\approx2+t/2=2\left(1+\dfrac{t}{4}\right)$, giving $\sqrt{1+\sqrt{1+t}}\approx\sqrt2\left(1+\dfrac{t}{8}\right)$. So the expression $\approx\sqrt2\cdot\dfrac{t}{8t}=\dfrac{1}{4\sqrt2}$, and the limit exists.

Solution: Since $\sin nx\approx nx$, the expression $\approx\dfrac{n[(a-n)nx-\tan x]}{x}$. Using $\tan x\approx x+x^3/3$, this tends to $n[(a-n)n-1]$. Setting this equal to $0$ gives $(a-n)n=1\Rightarrow a=n+\dfrac1n$.

Solution: Each term telescopes: $\tan^{-1}\dfrac{1}{1+(x+j)(x+j-1)}=\tan^{-1}(x+j)-\tan^{-1}(x+j-1)$, so $f_n(x)=\tan^{-1}(x+n)-\tan^{-1}x$. At $x=0$, $f_j(0)=\tan^{-1}j$, so $\tan(f_j(0))=j$; hence $\sum_{j=1}^5j^2=55$, proving (a). Also $f_j'(0)=\dfrac{1}{1+j^2}-1$, so $(1+f_j'(0))\sec^2(f_j(0))=\dfrac{1}{1+j^2}(1+j^2)=1$, and summing $j=1$ to $10$ gives $10$, proving (b). As $x\to\infty$, $f_n(x)\to0$, so $\tan(f_n(x))\to0$ (not $1/n$, so (c) is false) and $\sec^2(f_n(x))\to\sec^2(0)=1$, proving (d).

Solution: Expand $\sqrt{a^2-x^2}=a-\dfrac{x^2}{2a}-\dfrac{x^4}{8a^3}-\cdots$, so $a-\sqrt{a^2-x^2}-\dfrac{x^2}{4}=\left(\dfrac{1}{2a}-\dfrac14\right)x^2+\dfrac{x^4}{8a^3}+\cdots$. For $L$ to be finite (dividing by $x^4$), the $x^2$ coefficient must vanish: $\dfrac{1}{2a}=\dfrac14\Rightarrow a=2$. Then $L=\dfrac{1}{8a^3}=\dfrac{1}{64}$.

Solution: $\sin(\beta x)\approx\beta x$, so the numerator $\approx\beta x^3$. The denominator $\alpha x-\sin x\approx(\alpha-1)x+\dfrac{x^3}{6}$. For the limit to be finite and non-zero, we need $\alpha=1$, leaving denominator $\approx x^3/6$. Then the limit becomes $\dfrac{\beta x^3}{x^3/6}=6\beta=1\Rightarrow\beta=\dfrac16$. So $6(\alpha+\beta)=6\left(1+\dfrac16\right)=7$.

Solution: $\cos(\alpha^n)\approx1-\dfrac{\alpha^{2n}}{2}$, so $e^{\cos(\alpha^n)}\approx e\cdot e^{-\alpha^{2n}/2}\approx e\left(1-\dfrac{\alpha^{2n}}{2}\right)$, giving $e^{\cos(\alpha^n)}-e\approx-\dfrac{e\alpha^{2n}}{2}$. Dividing by $\alpha^m$ gives $-\dfrac{e}{2}\alpha^{2n-m}$, which is finite and equal to $-\dfrac{e}{2}$ only if $2n-m=0$, i.e. $m=2n$. So $\dfrac{m}{n}=2$.

Solution: As $x\to0$, $\dfrac{f(3+x)-f(3)}{x}\to f'(3)$, so the numerator $\to e^{f'(3)}$; similarly the denominator $\to e^{f'(2)}$. The limit is $e^{f'(3)-f'(2)}=e^0=1$.

Solution: Put $x=\sin\theta$; as $x\to1^-,\ \theta\to\pi/2^-$. With $t=\pi/2-\theta\to0^+$, $\sqrt{\pi}-\sqrt{2\sin^{-1}x}=\sqrt\pi-\sqrt{\pi-2t}\approx t/\sqrt\pi$ and $\sqrt{1-x}=\sqrt{1-\cos t}\approx t/\sqrt2$, so the ratio $\to\sqrt{2/\pi}$.

Solution: For $x\to0^+$, $[x]=0$, so $|x|-\sin(x[x])=x$ and the limit becomes $\pi-1$. For $x\to0^-$, $[x]=-1$, so $|x|-\sin(x[x])=-x-\sin(-x)=-x+\sin x$, which is $O(x^3)$, making that piece vanish and the limit becomes $\pi$. Since the two one-sided limits disagree, the limit does not exist.

Solution: As $x\to1^-$, $0<1-x<1$ so $[1-x]=0$ throughout, which makes $\sin\!\left(\frac{\pi}{2}[1-x]\right)=\sin 0=0$ identically for all such $x$, forcing the whole expression to be $0$ near $x=1^-$; hence the limit is $0$.

Solution: For $x\to1^-$, $|1-x|=1-x$, so the bracket simplifies to a value bounded away from $0$, while $\cos\!\left(\dfrac{1}{1-x}\right)$ oscillates between $-1$ and $1$ infinitely often as $x\to1^-$; hence the limit does not exist.

Solution: As $a\to0^+$, $\sqrt[n]{1+a}-1\sim a/n$, so the coefficients scale in the ratio $\frac13:\frac12:\frac16=2:3:1$, giving the limiting equation $2x^2+3x+1=0$, whose roots are $-1$ and $-\dfrac12$.

Solution: With $x=1+h$, $h\to0$: using $\sin h\sim h$, the base $\to\dfrac{a-1}{2}$ and the exponent $\dfrac{1-x}{1-\sqrt x}\to2$. The equation becomes $\left(\dfrac{a-1}{2}\right)^2=\dfrac14$, so $a-1=\pm1$, giving $a=0$ or $a=2$. The largest non-negative integer value is $2$.

Solution: Writing $x+\tfrac nk=\tfrac nk\left(1+\tfrac{kx}{n}\right)$ and $x^2+\tfrac{n^2}{k^2}=\tfrac{n^2}{k^2}\left(1+\tfrac{k^2x^2}{n^2}\right)$ and simplifying using $n!=\prod k$, $f(x)=\displaystyle\lim_{n\to\infty}\left[\prod_{k=1}^{n}\dfrac{1+kx/n}{1+(kx/n)^2}\right]^{x/n}$. Taking logs and converting the Riemann sum to an integral gives $\ln f(x)=x\displaystyle\int_0^1\ln\dfrac{1+xt}{1+x^2t^2}\,dt=(1+x)\ln(1+x)-x\ln(1+x^2)-2\tan^{-1}x+x$. Differentiating, $\dfrac{f'(x)}{f(x)}=\ln\dfrac{1+x}{1+x^2}$. For $0<x<1$, $1+x>1+x^2$, so $f'(x)/f(x)>0$ there, i.e. $f$ is increasing on $(0,1)$; hence $f(1/2)<f(1)$ (so (a) is false) while $f(1/3)<f(2/3)$ (so (b) is true). At $x=2$, $f'(2)/f(2)=\ln(3/5)<0$ and $f(2)>0$, so $f'(2)<0$ (c) is true. At $x=3$, $f'(3)/f(3)=\ln(2/5)<\ln(3/5)=f'(2)/f(2)$, so (d) is false. Hence (b) and (c) are correct.

Solution: $\ln y_n=\dfrac1n\displaystyle\sum_{k=1}^{n}\ln\left(1+\dfrac kn\right)\to\int_0^1\ln(1+x)\,dx=\Big[(1+x)\ln(1+x)-x\Big]_0^1=2\ln2-1=\ln4-1$. So $L=e^{\ln4-1}=\dfrac4e\approx1.4715$, giving $[L]=1$.

Solution: Because of $\sqrt{x}$, $f$ is defined only for $x \geq 0$, so at $x = 0$ only a right-hand derivative can be examined: $f'(0^+) = \lim_{x\to 0^+} \dfrac{f(x)-f(0)}{x} = \lim_{x\to 0^+}\left(\sqrt{x}-\sqrt{x+1}\right) = -1$. Since $f$ is not even defined for $x<0$, a genuine (two-sided) derivative cannot exist at $x=0$, so $f$ is not differentiable there.